Problem: $h(x)=2x^4+6x^3-12x^2+8$. On which intervals is the graph of $h$ concave up? Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-\infty,-2\right)$ and $\left(\dfrac12,\infty\right)$ (Choice B) B $\left(-2,\dfrac12\right)$ only (Choice C) C $\left(-\infty,-\dfrac12\right)$ and $\left(2,\infty\right)$ (Choice D) D $\left(-6,-4\right)$ only
Explanation: We can analyze the intervals where $h$ is concave up/down by looking for the intervals where its second derivative $h''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $h'$, we are analyzing $h''$. The second derivative of $h$ is $h''(x)=12(x+2)(2x-1)$. $h''(x)=0$ for $x=-2,\dfrac12$. Since $h''$ is a polynomial, it's defined for all real numbers. Therefore, our points of interest are $x=-2$ and $x=\dfrac12$. Our points of interest divide the number line into three intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $(-\infty, \llap{-}2)$ $\left( \llap{-}2,\frac12\right)$ $\left(\frac12,\infty\right)$ Let's evaluate $h''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h''(x)$ Verdict $(-\infty,-2)$ $x=-3$ $h''(-3)=84>0$ $h$ is concave up $\cup$ $\left(-2,\dfrac12\right)$ $x=0$ $h''(0)=-24<0$ $h$ is concave down $\cap$ $\left(\dfrac12,\infty\right)$ $x=1$ $h''(1)=36>0$ $h$ is concave up $\cup$ In conclusion, the graph of $h$ is concave up over the intervals $\left(-\infty,-2\right)$ and $\left(\dfrac12,\infty\right)$.